PM is the median of ∆PQR then prove that PQ² + PR² = ( PM² + QM² )
anwer right i need right answer. .it is board level question so. .plz..right. .I will mark as brainliast
anwer right i need right answer. .it is board level question so. .plz..right. .I will mark as brainliast
Answers & Comments
Answer:
In ∆ PQR, PR²-PQ²= QR² & QM ⊥ PR
To Prove: QM² = PM × MR
Proof:
Since, PR² - PQ²= QR²
PR² = PQ² + QR²
So, ∆ PQR is a right angled triangle at Q.
In ∆ QMR & ∆PMQ
∠QMR = ∠PMQ [ Each 90°]
∠MQR = ∠QPM [each equal to (90°- ∠R)]
∆ QMR ~ ∆PMQ [ by AA similarity criterion]
By property of area of similar triangles,
ar(∆ QMR ) / ar(∆PMQ)= QM²/PM²
1/2× MR × QM / ½ × PM ×QM = QM²/PM²
[ Area of triangle= ½ base × height]
MR / PM = QM²/PM²
QM² × PM = PM² × MR
QM² =( PM² × MR)/ PM
QM² = PM × MR
Answer:
In ∆ PQR, PR²-PQ²= QR² & QM ⊥ PR
To Prove: QM² = PM × MR
Proof:
Since, PR² - PQ²= QR²
PR² = PQ² + QR²
So, ∆ PQR is a right angled triangle at Q.
In ∆ QMR & ∆PMQ
∠QMR = ∠PMQ [ Each 90°]
∠MQR = ∠QPM [each equal to (90°- ∠R)]
∆ QMR ~ ∆PMQ [ by AA similarity criterion]
By property of area of similar triangles,
ar(∆ QMR ) / ar(∆PMQ)= QM²/PM²
1/2× MR × QM / ½ × PM ×QM = QM²/PM²
[ Area of triangle= ½ base × height]
MR / PM = QM²/PM²
QM² × PM = PM² × MR
QM² =( PM² × MR)/ PM
QM² = PM × MR
Step-by-step explanation: